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How can I calculate the following limit epsilon-delta definition?

$$\lim_{x \to 0} \left(\frac{\sin(ax)}{x}\right)$$

Edited the equation, sorry...

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Did you type the problem correctly? Is it supposed to be Sin(ax) / x? – Amzoti Nov 28 '12 at 16:18
I wonder: are you taking $\lim_{y\rightarrow 0}\sin(y)/y=1$ as granted? There is a way to solve this without epsilon and delta, if so... – rschwieb Nov 28 '12 at 16:26
@rschwieb I proved that earlier using epsilon delta, so yes. – lazyCrab Nov 28 '12 at 16:28
This kind of questions are odd: if you want an $\,\epsilon-\delta\,$ proof then it is because you already know, or at least heavily suspect, what the limit is...and if you already know/suspect this, it is because you can evaluate the limit by other means, so... – DonAntonio Nov 28 '12 at 17:45
@lazyCrab, you proved $\,\frac{\sin y}{y}\xrightarrow [y\to 0]{}1\,$ by epsilon delta?! Would you mind to post your proof, please? – DonAntonio Nov 28 '12 at 17:46
up vote 2 down vote accepted

you can rewrite $\dfrac{\sin (ax)}{x} = a \dfrac{\sin (ax)}{ax}$, then take limits, as suggested by @rscwieb in the comments

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Can you please be more specific? – lazyCrab Nov 28 '12 at 16:56
so $\lim_{x \rightarrow 0}\dfrac{\sin (ax)}{x} = a \lim_{x \rightarrow 0} \dfrac{\sin (ax)}{ax} = a \cdot 1 = a$? I was hoping you'll get it as it looks like homework – anegligibleperson Nov 28 '12 at 17:00
Thanks, just wanted to be sure. Though this does not use epsilon delta at all. – lazyCrab Nov 28 '12 at 17:17

This is easy if you know the power series of the sine function.

$$\sin(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} \dots$$

Then $$\sin(ax) = a x - \frac{a^3 x^3}{6} + \frac{a^5 x^5}{120} \dots$$

and $\frac{\sin(ax)}{x} = a - \frac{a^3 }{6}x^2 + \frac{a^5 }{120}x^4 \dots$ for all $x \neq 0$. Since power series are continuous you can find the limit for $x \rightarrow 0$ by simply setting $x=0$ in this expression, so the limit is $a$.

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Sorry, haven't "formally" learnt it yet. – lazyCrab Nov 28 '12 at 16:56

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